Why is this series convergent

Here are the pictures of the problem I am talking about. http://www.flickr.com/photos/venom104 . I get that the first term is a geometric series and is convergent because r < 1. What I don't get, is the term after using partial fraction decomposition, about how that series is convergent. They show the steps, but my book says..."Theorem 8...sigma(n=1,inf) a(n) - b(n) = sigma(n=1,inf) a(n) - sigma(n=1,inf) b(n).

So if that theorem is true, then the series can be split up into sigma(n=1,inf) 1 / n - sigma(n=1,inf) 1 / n+1). If is split up that way, by definition the series of 1/n is DIVERGENT because it is a HARMONIC series.

Is the book (the book has the same answer) and the posted solution here wrong?

Careful with the hypotheses of the theorem!

If both Σ(n = 1 to ∞) a(n) and Σ(n = 1 to ∞) b(n) are both convergent, then
Σ(n = 1 to ∞) (a(n) - b(n)) = Σ(n = 1 to ∞) a(n) - Σ(n = 1 to ∞) b(n).

You're trying to use a converse to a theorem for which the converse makes no sense.

That is, we can't use this theorem to conclude that
Σ(n = 1 to ∞) (1/n - 1/(n-1)) = Σ(n = 1 to ∞) 1/n - Σ(n = 1 to ∞) 1/(n-1),
because the individual series on the right side don't even converge.

Since they don't fit the requirements of the theorem, we can't apply it.

I hope this helps!

The reason B is convergent is the each term in the summation is composed of +(1st number) - (second number) but the first term in the next set (when you increment the summand) has the same magnitude but opposite sign as the second term in the previous one so they cancel out.

for n=1 => +1/1-1/(1+1)=+1-1/2
for n=2 => +1/2-1/(2+1)=+1/2-1/3
.......
you see that the -1/2 and +1/2 cancel each other out so you will be left with +1-1/(infinity+1) which is shown in the next line of the picture